# Help with 70-291 Question

Member Posts: 5 ■□□□□□□□□□
I ran accross this question from the MCSA/MCSE 70-291 Self-Paced Training Kit. I am not sure why the correct answer is correct. Can anyone give me an in depth explanation to how they arrived at this answer on this question?

You work for a large organization that has leased the Class B
address 131.188.0.0. It uses VLSM. The network is divided into 6
backbone subnets, each with 8190 host addresses. These hosts are
then subnetted as required. The 131.188.96.0/19 subnet is divided
into 62 equal-sized subnets. On which subnet would you find host
131.188.97.140?

1. 131.188.97.128/25 <Correct>
2. 131.188.96.0/19
3. 131.188.97.0/25
4. 131.188.97.0/24

Explanation :
The host address range for this subnet is 131.188.97.129 through
131.188.97.254. It is also the most specific subnet in the path.
Host 131.188.97.140 is found on this subnet.

Objective:
Implementing, Managing, and Maintaining IP Addressing
Sub Objective(s):
Configure TCP/IP addressing on a server computer

:

• Member Posts: 179
the "/25" indicates the first 25 bits are used to identify the unique network
(11111111.11111111.11111111.10000000)
leaving the remaining bits to identify the specific host.
(The remaining bits can add up to 1111111 = 127)

so if you take the .128 + 127 = 255 -1(255 is used as broadcast) = 254
Meaning that all valid hosts need to be in the range of 131.188.97.129 to 131.188.97.254

Kind regards.
Eastp